Question

QUESTION 1 [25]

A student applies for University admission in China (A) and German (B) and South Africa (C) The probability that he is admitted in China is 0.213 the probability of being admitted is South Africa is 0.275. The probability of being admitted in all countries is 0.185. The probability of being admitted in at least country is 0.495. The probability of being admitted to China and South Africa is 0.245 while the probability of being admitted to China and German is 0.203. (Show all your work)

a. What is the probability be admitted in Germany P(B)? [6]

b. Are events A, B and C mutually exclusive? Explain, using probabilities.

Why or why not? Explain. [3]

c. Are the two events A, B and C independent? Explain, using probabilities. [3]

d. What is the probability of being admitted in Germany given that a student has been admitted in China? [4]

e.What is the probability of being admitted in China given that a student has been admitted in South Africa? [4]

f. What is the probability of being admitted in German and South Africa if P(B|C) is 0.975 [5]

Answer 1

given

p(A)=0.213

P(C)=0.275

P(AUBUC)=0.495

P(A∩C)=0.245

P(A∩B)=0.203

P(A∩B∩C)=0.185

a) p(B)=P(AUBUC)=P(A)-P(C)+P(A∩B)+P(A∩C)+P(B∩C)-P(A∩B∩C)

P(B/C)=0.975

P(B∩C)=0.975*0.275=0.268

P(B)=0.495-0.213-0.275+0.203+0.245+0.268-0.185

=0.538

b)the three events A,B and c are not mutually exclusive

c) their intersection is not zero so A,B,C are independent

d) we have to find P(B/A)

P(B/A)=P(A∩B)/P(A)

=0.203/0.213

=0.953

e)p(A/C)=P(A∩C)/P(C)

= 0.245/0.275

=0.891

f) p(B∩C)=0.975*0.275=0.268